Session 19
Frequency Response of
Transistor Amplifiers
Electronic Components and Circuits
Pablo Acedo
www.uc3m.es/portal/page/portal/dpto_tecnologia_electronica/Personal/PabloAcedo
Frequency Response of Transistor
Amplifiers
INDEX
•
•
•
•
Fundamentals and tools for frequency response
analysis of transistor amplifier circuits.
BJT and FET small signal high-frequency models.
High-Frequency response analysis of amplifier circuits.
Open-circuit time constants method. Examples.
Low-Frequency response analysis of amplifier circuits.
Short-circuit time constants method. Example.
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CCE - Session 19
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Fundamentals
Low-pass RC Circuit
20log|Vo(jω)/Vg(jω) | (dB)
0dB
-20dB
ω=2πf
V
1
T (s) = o =
Vg 1 + ( s / ω0 )
T ( jω ) =
0º
1 + (ω / ω0 )2
-45º
φ (ω ) = − tan (ω / ω0 )
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ωo
10ωo
ωo/10
ωo
10ωo
φ (ω)
1
−1
ωo/10
ω=2πf
-90º
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Fundamentals
High-pass RC Circuit
20log|Vo(jω)/Vg(jω) | (dB)
0dB
-20dB
Vo
s
T (s) =
=
Vg s + ω0
ωo/10
90º
1 + (ω0 / ω ) 2
45º
φ (ω ) = tan −1 (ω0 / ω )
0º
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10ωo
ωo/10 ωo
10ωo
ω=2πf
φ (ω)
1
T ( jω ) =
ωo
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ω=2πf
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Fundamentals
Bode Plot (Amplitude and Phase)
20log|Vo(jω)/Vg(jω) | (dB)
20log(Ao) dB
20dB
ω1/10
φ (ω)
90º
45º
0º
-45º
-90º
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ω1 10ω1
ω2/10 ω2 10ω2
10s
T ( s) =
1 + s / 102 * 1 + s / 105
(
)(
ω=2πf
)
ω2/10 ω2 10ω2
ω1/10 ω1 10ω1
CCE - Session 19
ω=2πf
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Fundamentals: The three Frequency Bands
20log|Av (jω)|
(dB)
Midband
Lowfrequency
band
Ci, Co, Ce
20log(Ao) dB
fci
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Highfrequency
band
Cπ, Cµ
Bandwidth
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fcs
f (Hz)
6
BJT high-frequency small-signal model
Unity-gain
bandwidth
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gm
fT =
2π (Cπ + Cµ )
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FET high-frequency small-signal model
Unity-gain
bandwidth
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gm
fT =
2π (C gs + C gd )
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High-frequency response analysis of amplifier
circuits.
•
•
•
High frequency response of transistor amplifier circuits
is fixed by the internal capacitances of the devices and
the associated time constants.
In general, we will assume that the high-frequency
response is fixed by a DOMINANT POLE. This way, the
high-frequency analysis is reduced to the calculation of
such dominant pole.
The dominant pole frequency calculation will be
performed using the Open-circuit time constants
method.
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CCE - Session 19
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Open-circuit time constants method.
1
f cs (3dB) =
2π
1
0
∑ Ri Ci
i
Where:
Ci are all the capacitances that influence in the highfrequency response: active devices internal
capacitances, or small capacitors (with low-pass
characteristics), introduced in the circuit to control the
high-frequency response.
R0i is the resistance seen by each of the capacitors
when we reduce all other capacitances to zero (Open
circuit)
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Open-circuit time constants method.
Example (I)
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Open-circuit time constants method.
Example (II)
Rµ0 = RB // rπ // Rg (1 + g m RC // RL ) + RC // RL
Rπ0 = RB // rπ // Rg
1
1
fcs =
2π Cπ (rπ // RB // Rg ) + Cµ (RB // rπ // Rg (1+ gmRC // RL ) + (RC // RL ))
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Miller’s Theorem. Example (I)
Cmu (1-K)
K = − gm RC // RL
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Miller’s Theorem. Example (II)
Miller’s Theorem vs Dominant Pole Calculation
1
1
fcsM =
2π (rπ // RB // Rg ) [Cπ + Cµ (1+ gmRC // RL )]
1
1
fcs =
2π Cπ (rπ // RB // Rg ) + Cµ (RB // rπ // Rg (1+ gmRC // RL ) + RC // RL )
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CCE - Session 19
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Low-frequency response analysis of amplifier
circuits.
•
•
•
Low-frequency response of transistor circuits is fixed by
the coupling and bypass capacitors and the associated
time constants.
In general, we will assume that the low-frequency
response is fixed by a DOMINANT POLE. This way, the
low-frequency analysis is reduced to the calculation of
such dominant pole.
The dominant pole frequency calculation will be
performed using the Short-circuit time constants
method.
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CCE - Session 19
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Short-circuit time constants method.
1
f ci (3dB) =
2π
1
∑ R ∞C
i
i
i
Where:
Ci are all the coupling and bypass capacitors of the
circuit.
R∞i is the resistance seen by each of the capacitors
when we assume all other capacitances are infinity
(Short circuit)
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Short-circuit time constants method.
EXAMPLE (I)
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Short-circuit time constants method.
EXAMPLE (II)
∞
RCi
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= Rg + RB // rπ
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Short-circuit time constants method.
EXAMPLE (III)
∞
RE
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= RE //
rπ + ( Rg // RB )
1 + β0
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Short-circuit time constants method.
EXAMPLE (IV)
∞
Co
R = RC + RL
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Short-circuit time constants method.
EXAMPLE (V)




1
1
1
1

fci =
+
+
rπ + (Rg // RB )  Co (RC + RL )
2π Ci (Rg + RB // rπ )


CE  RE //


1+ β0




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Proposed Exercise
+Vdd = 15 V
RS = 560Ω
RG = 1 MΩ
Rg = 50 Ω
Transistor:
IDSS= 10 mA
RD = 5,6 KΩ
RL=10 KΩ
Ci = 10 µF
C0=10 µF
VP = -2 V
Cgd=0.36 pF
Cgs=1 pF
ID = IDSS · (1-VGS/VP)2
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